2. 4/5 points | Previous Answers OSColPhys1 16.P.019.Tutorial.WA. My Notes Ask Y
ID: 1353021 • Letter: 2
Question
2. 4/5 points | Previous Answers OSColPhys1 16.P.019.Tutorial.WA. My Notes Ask Your Teacher A mass of 0.24 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by x(t) = (0.48 m)cos[(18 rad/s)t]. xt)-(0.48 m)cost(18 rad/s)t]. Determine the following. Determine the following (a) amplitude of oscillation for the oscillating mass 0.48 (b) force constant for the spring 77.76 (c) position of the mass after it has been oscillating for one half a period 0.48 (d) position of the mass one-third of a period after it has been released N/mm 0.24 (e) time it takes the mass to get to the position x-0.10 m after it has been released (e) time it takes the mass to get to the position x =-0.10 m after it has been releasedExplanation / Answer
SOLUTION:
In general, x(t) = A cos(t - ), where A is the amplitude, is the angular frequency, and is some phase shift.
(a)
0.48 m
(b)
For the spring-mass system,
² = k/m
(18 rad/s)² = k / (0.24 kg)
k = 77.76 N/m
(c)
x(0.48 s) = (0.50 m)cos[(18 rad/s)(0.5 s)]
x(0.48 s) = 0.49 m
(d)
x(t) = (0.48 m)cos[2/3]
x(t) = 0.24 m
(e)
-0.10 m = (0.48 m)cos[(18 rad/s)(t)]
cos(18t) = -0.20
18t = 112.81
18t=1.95
t = 0.10 s
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